Question: Let $f(x)=\dfrac{1}{x^7}$. $f'(x)=$
Explanation: The strategy We can first rewrite $f(x)$ as a negative power of $x$. Then, the derivative of $f$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting the fraction as a negative power $f(x)=\dfrac{1}{x^7}=x^{-7}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(x^{-7}\right) \\\\ &=-7x^{-7-1} \gray{\text{The power rule}} \\\\ &=-7x^{-8} \end{aligned}$ In conclusion, we found that $f'(x)=-7x^{-8}$. This can also be written as $-\dfrac{7}{x^8}$ (all equivalent forms are accepted).